Question

The magnetic field of a plane electromagnetic wave is given by $\overrightarrow{B}=B_0\hat{i}\left[cos\left(kz-\omega t\right)\right]+B_1\hat{j}cos\left(kz+\omega t\right)$ , where $B_0=3\times 10^{-5}T$ and $B_1=2\times 10^{-6}T$ . The $RMS$ value of the force experienced by a stationary charge $Q=10^{-4}C$ at $z=0$ is closest to

• A. $0.1N$
• B. $0.9N$
• C. $3\times 10^{-2}N$
• D. $0.6N$

Answer 1

Answer: (D)

$\overrightarrow{B}=B_0\hat{i}\left[cos\left(kz-\omega t\right)\right]+B_1\hat{j}cos\left(kz+\omega t\right)$
$B_0=3\times 10^{-5}T\&B_1=2\times 10^{-6}T$
Electric filed associated with it is,
$\overrightarrow{E}=-B_{0}c\hat{j}cos\left(kz-\omega t\right)-B_{1}c\hat{i}cos\left(kz+\omega t\right)$
Here, $c$ is the speed of light in vacuum.
At $z=0,$
Here $c$ is speed of light in vacuum.
$\begin{array}{l}=\sqrt{\frac{{\left(10^{-4}\times 3\times 10^{-5}\times 3\times 10^8\right)}^2+{\left(10^{-4}\times 2\times 10^{-6}\times 3\times 10^8\right)}^2}{2}}\\ =\sqrt{\frac{0.81+0.0036}{2}}=0.6N\end{array}$